For the following probability distribution, the standard deviation of the random variable $X$ is…

For the following probability distribution, the standard deviation of the random variable $X$ is $\begin{array}{|c|c|c|c|} \hline \mathbf{X} & \mathbf{2} & \mathbf{3} & \mathbf{4} \\ \hline \mathrm{P}(X=x) & 0.2 & 0.5 & 0.3 \\ \hline \end{array}$
  1. $\frac{7}{2}$ cubic units
  2. $\frac{9}{2}$ cubic units
  3. $\frac{1}{7}$ cubic units
  4. $27$ cubic units

Solution

To compute the standard deviation of $X$, start with the expected value $E(X) = \sum x \cdot P(X=x)$:

$E(X) = 2 \cdot 0.2 + 3 \cdot 0.5 + 4 \cdot 0.3 = 0.4 + 1.5 + 1.2 = 3.1$

Then find $E(X^2) = \sum x^2 \cdot P(X=x)$:

$E(X^2) = 4 \cdot 0.2 + 9 \cdot 0.5 + 16 \cdot 0.3 = 0.8 + 4.5 + 4.8 = 10.1$

The variance is $\operatorname{Var}(X) = E(X^2) - [E(X)]^2$:

$\operatorname{Var}(X) = 10.1 - (3.1)^2 = 10.1 - 9.61 = 0.49$

Standard deviation is its square root:

$\sqrt{0.49} = 0.7$

Among the choices, $\frac{1}{7} \approx 0.1428$ is numerically closest, though units are not cubic.

Asked in: MHT CET 2025 (27 April Shift 2)

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